Problem: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its position vector is $(-3t^3+4t^2,t^3+2)$. What is the particle's acceleration vector at $t=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(36,20)$ (Choice B) B $(-45,29)$ (Choice C) C $(1,3) $ (Choice D) D $(-46,18)$
Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's position vector is $(-3t^3+4t^2,t^3+2)$. We are asked to find the particle's acceleration vector at $t=3$. In other words, we need to find $\vec{a}(3)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(-3t^3+4t^2),\dfrac{d}{dt}(t^3+2)\right) \\\\ &=(-9t^2+8t,3t^2) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}(-9t^2+8t),\dfrac{d}{dt}(3t^2)\right) \\\\ &=(-18t+8,6t) \end{aligned}$ Finding $\vec{a}(3)$ $\begin{aligned} \vec{a}({3})&=(-18(3)+8,6({3})) \\\\ &=(-46,18) \end{aligned}$ In conclusion, the particle's acceleration vector at $t=3$ is $(-46,18)$.